Analysis Recap

Metric Spaces

Definition: A metric on a set $X$ is a function $d:X\times X\to \mathbb{R}$ that satisfies,

1. $d(x,y)\ge 0$ and $d(x,y)=0⟺x=y$
2. $d(x,y)=d(y,x)$
3. $d(x,z)\le d(x,y)+d(y,z)$
   for all $x,y,z\in X$

A metric space $(X,d)$ is a set equipped with a metric, although generally the metric of a space is inferred from context

Definition: A sequence $\{x_k{\}}_{k\in \mathbb{N}}$ in $X$ is said to converge to $x\in X$ (its limit) if for every $ϵ>0$ there exists an $N\in \mathbb{N}$ such that $d(x_k,x)<ϵ$ for all $k\ge N$. If this is the case, the sequence converges.

${\lim }_{k\to \infty }{x}_k=x$ is the notation for convergence introduced in calculus

If a limit of a sequence exists then it is unique

Definition: A sequence $\{x_k{\}}_{k\in \mathbb{N}}$ with $x_k\in X$ is a Cauchy sequence if for every $ϵ>0$ there exists an $N\in \mathbb{N}$ such that if $k,l\ge N$ then $d(x_k,x_l)<ϵ$.

The idea captured by a Cauchy sequence is that the elements in the tail get closer and closer to each other

Every convergent sequence is a Cauchy sequence but not vice versa (this can happen with stranger metric spaces)

Definition: A metric space $(X,d)$ is complete if every Cauchy sequence converges

Theorem: $(\mathbb{R},d)$, $({\mathbb{R}}^n,d_{\text{eucl}})$, $({\mathbb{R}}^n,d_{\text{man}})$ are complete metric spaces

A sequence is bounded if there exists an $x\in X$ and $M\in \mathbb{R}$ such that $d(x,x_k)\le M$ for all $k$

Bolzano-Weierstrass: A sequence in ${\mathbb{R}}^n$ has the property that every bounded sequence contains a Cauchy subsequence (and as such a convergent subsequence)

Convergence

We restrict ourselves to $\mathbb{R}$ or $\mathbb{C}$ at this point

Given a sequence $\{x_k{\}}_{k\in \mathbb{N}}$, we define the partial series $S_n={\sum }_{k=1}^n{x}_k$, which converges if $\{S_n{\}}_{n\in \mathbb{N}}$ converges

A series converges absolutely if ${\sum }_{k=1}^{\infty }|x_k|$ converges

Theorem: If a series converges absolutely then it converges

One might expect that a convergent series under any bijection $S_{\varphi }={\sum }_{k=1}^{\infty }{x}_{\varphi (k)}$ will converge, but actually multiple situations can occur,

• $S$ and $S_{\varphi }$ both converge and $S=S_{\varphi }$
• $S$ converges, but $S_{\varphi }$ does not
• $S$ and $S_{\varphi }$ both converge but $S\ne S_{\varphi }$
  However, these are only true if $S$ does not converge absolutely

Theorem: Let $\varphi :\mathbb{N}\to \mathbb{N}$ be a bijection, if $S={\sum }_{k=1}^{\infty }{x}_k$ converges absolutely then $S_{\varphi }={\sum }_{k=1}^{\infty }{x}_{\varphi (k)}$ also converges absolutely and $S=S_{\varphi }$

Non-absolute convergence can be very deceptive, for example given a real number $x$, you can always find a bijection of $\mathbb{N}$ such so that the rearranged serries converges to $x$ in this case (Reimann’s rearrangement theorem)

We say that a bi-infinite series converges absolutely if the partial symmetric sums $s_n={\sum }_{k=-n}^n{x}_k$ converges absolutely, in which case both $S_{\ge 0}$ and $S_{<0}$ converge absolutely and $S=S_{\ge 0}+S_{<0}$

Continuous Functions

Definition: A function $f:X\to Y$ between $(X,d_X)$ and $(Y,d_Y)$ is continuous at $x\in X$ if for every $ϵ>0$ there exists a $\delta >0$ such that if $d_X(x,y)<\delta$ then $d_Y(f(x),f(y))<ϵ$. If $f$ is continuous at every $x\in X$, then $f$ is called continuous.

We like continuous functions :)

Theorem: Suppose $f:X\to Y$ is continuous and suppose a sequence $x_k$ converges to $x$, then $f(x_k)$ converges to $f(x)$

${\lim }_{k\to \infty }f(x_k)=f({\lim }_{k\to \infty }{x}_k)$

The converse is that if for every sequence $\{x_k{\}}_{k\in \mathbb{N}}$ which converges to $x$ we have that $\{f(x_k){\}}_{k\in \mathbb{N}}$ converges to $f(x)$ then $f$ is continuous at $x$

Sequences of Functions

Let $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and $I\subset \mathbb{R}$ be an interval
We define $f:I\to \mathbb{F}$ and $f_k:I\to \mathbb{F}$

Definition: $f_k$ converges point-wise to $f$ if for every $x\in I$, $\{f_k(x){\}}_{k\in \mathbb{N}}$ in $\mathbb{F}$ converges to $f(x)$

This definition has a couple annoyances

Example: Let $f_k:[0,1]\to \mathbb{R}$ be given by $f_k(x)=x^k$
$f_k$ converges point-wise to the function
$f(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array}$
$f_k$ is continuous but the point-wise limit is not

This also behaves badly with Reimann integrals

Definition: $f_k$ converges uniformly to $f$ if for every $ϵ>0$ there exists an $N$ such that ${\text{sup}}_{x\in I}d(f_k(x),f(x))<ϵ$ for all $k\ge N$

In other words, ${\text{sup}}_{x\in I}d(f_k(x),f(x))\to 0$

It figures that uniform convergence is a stronger condition than point-wise convergence

Theorem: Suppose $f_k$ is a sequence of continuous functions and suppose that the sequence converges uniformly to $f$, then $f$ is continuous

Theorem: Let $a<b$ and suppose each term in $f_k:[a,b]\to \mathbb{F}$ is Reimann integrable. Suppose that $f_k$ converges uniformly to $f$, then $f$ is Reimann integrable and ${\lim }_{k\to \infty }{\int }_a^b{f}_k(x)dx={\int }_a^{b}f(x)dx$

The Uniform Metric

Can we reformulate uniform convergence in terms of a metric on the space of all functions?

$d_{\infty }(f,g)={\text{sup}}_{x\in I}d(f(x),g(x))$

One problem with this metric is that we can’t guarantee it is bounded, so we must specify this requirement

$B(I)=\{f:I\to \mathbb{F}\mid f\text{ is bounded}\}$
$(B,d_{\infty })$ is a metric space
$(B(I),d_{\infty })$ is complete

Theorem: The metric space $(C([a,b]),d_{\infty })$ is complete

We denote $\text{RI}([a,b])\{f:[a,b]\to \mathbb{F}\mid f\text{ is Reimann integrable}\}$
This is bounded definitionally, so we can apply our uniform distance operator

Theorem: $(\text{RI}([a,b]),d_{\infty })$ is a complete metric space. Let $\int :\text{RI}([a,b])\to \mathbb{R}$ assign $f\in \text{RI}([a,b])$ to $\int (f)={\int }_a^{b}f(x)dx$. Then $\int$ is a continuous mapping.

We can also define a metric that works for all functions like ${\bar{d}}_{\infty }(f,g)=\{\begin{array}{ll}{\text{sup}}_{x\in I}d(f(x),g(x))& {\text{if sup}}_{x\in I}d(f(x),g(x))<1\\ 1& \text{otherwise}\end{array}$

There is no metric that is equivalent to point-wise convergence

We denote $C^0(I)=C(I)$ the space of continuous functions and introduce $C^k(I)=\{f:I\to \mathbb{F}\mid f\text{ is differentiable and }{f}^{\prime }\in C^{k-1}(I)\}$ for $k\in N$

$C^k(I)$ are functions that are $k$ times differentiable with $k$-th derivatives continuous (note $C^{k+1}(I)\subset C^k(I)$)

We can define the smooth functions on $I$ by $C^{\infty }(I)={\bigcap }_{k\in N}^{}{C}^k(I)$